We have the finitely generated abelian group $$G:=\langle x_1,x_2,x_3: x_1+x_2+4x_3=x_1+4x_2+x_3=4x_1+x_2+x_3=0_G \rangle. $$ Let $A:=\langle f_1,f_2,f_3 \rangle$ be a free abelian group with rank $3$. If $H:=\langle f_1+f_2+4f_3,f_1+4f_2+f_3,4f_1+f_2+f_3\rangle$, we want to show that $$A/H\cong G.$$ We define the function $\phi:A\longrightarrow G,\ a=k_1f_1+k_2f_2+k_3f_3\mapsto \phi(a):=k_1x_1+k_2x_2+k_3x_3.$ And it's not difficult to see that this is a group epimorphism. But I faced a problem on the kernel:

We want to prove that $\ker{\phi}=H$.

$``\supseteq"$ We have that $\phi(f_1+f_2+4f_3)=x_1+x_2+4x_3=0_G,\ \phi(f_1+4f_2+f_3)=x_1+4x_2+x_3=0_G,\ \phi(4f_1+f_2+f_3)=4x_1+x_2+x_3=0_G \implies f_1+f_2+4f_3,f_1+4f_2+f_3,4f_1+f_2+f_3 \in \ker \phi \implies \langle f_1+f_2+4f_3,f_1+4f_2+f_3,4f_1+f_2+f_3 \rangle = H \subseteq \ker \phi.$

$``\subseteq "$ Let $z\in \ker \phi =\{ a\in A: \phi(a)=0_G \}$. Then, $z\in A\iff z=k_1f_1+k_2f_2+k_3f_3,\ k_i\in \Bbb{Z}$ and $k_1x_1+k_2x_2+k_3x_3=0_G$.


  1. How could we show now that $z\in H$?

  2. The group $$G=\{\mu_1x_1+\mu_2x_2+\mu_3x_3: \mu_1,\mu_2,\mu_3\in \Bbb{Z} \text{ and }x_1+x_2+4x_3=x_1+4x_2+x_3=4x_1+x_2+x_3=0_G \},$$ right?

Thank you.

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    $\begingroup$ If $k_1x_1+k_2x_2+k_3x_2=0_G$, what are the possible values of $k_1,k_2,k_3$ according to the presentationf of $G$? $\endgroup$ – Javi May 8 '18 at 20:38
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    $\begingroup$ How is the presented group defined? It's usual to take this quotient construction on the free group as the definition of group presentations. $\endgroup$ – Berci May 8 '18 at 20:42
  • $\begingroup$ @Javi $1$ or $4$? $\endgroup$ – Chris May 8 '18 at 20:52
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    $\begingroup$ Yes, $x_1,x_2,x_3$ are fixed. Your set definition in 2), especially the conditions therein with $x_1,x_2,x_3$ is not clear. In class, how the presentation was defined if not by the quotient? Maybe by a universal property? $\endgroup$ – Berci May 8 '18 at 22:02
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    $\begingroup$ It's usually defined by this very construction $A/H$. $\endgroup$ – Berci May 8 '18 at 22:12

In general, the group presentation $G=\langle x_i\,\vert\, \tau_j=0\rangle$, where $\tau_j$ are linear terms using variables $x_i$'s, means that $G$ is generated by the elements $x_i$ and exactly the given relations, and their consequences are satisfied in $G$. So that, whenever $\tau(.\!.x_i.\!.\!)=0$ in $G$ for a term $\tau$, it must be a consequence of the given relations.

Observing that the consequences in this case are just $\Bbb Z$-linear combinations, we can make a precise definition of the presented group as the quotient $F(.\!.x_i.\!.\!)\,/\,(.\!.a_j.\!.\!)$ where $F$ denotes the free Abelian group.

Said this, 1. requires no proof.

The presentation has the important universal property, similar to that of free groups, that whenever an Abelian group $Y$ is given with elements $y_i\in Y$ such that all $\tau_j(.\!.x_i.\!.\!)=0$ hold, there is a unique group homomorphism $f:G\to Y$ with $f(x_i)=y_i$.
It's also possible to define group presentation by this property.

For 2., your set notation is not clear: the elements $x_1,x_2,x_3$ don't vary, and they live inside the set you're about to define.

Note also that group presentations nicely generalize to any equationally defined class of algebraic structures.


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